Unfortunately, there is no cut and dry way to handle every related rates problem. There are some very common setups that you will see show up, but they are the type of problem that you have to be willing to play around with the pieces to find how they all come together. If you really want to master related rates problems, you have to do a bunch of them. The more of them you attempt, the more different models you will see, and the quicker you will start being able to make the connections. The problem you see on a test or the AP Calculus exam might be different in what the things in the problem are, but the method might match a technique from another problem you have done. If you are talking about a conical pile of sand in one problem you did, that problem’s technique might relate to the problem about a conical funnel with water because they are both most likely going to use the volume of a cone equation.
Here is the general method I follow when approaching any Related Rates problem.
Step 1: Read through the problem once, and sketch out a diagram of what is happening.
When you are first starting out with these, make that diagram big. Do not try to squeeze it into a little corner on your page. You are going to need the label that image with lots of data, and then try to connect that data. Draw big so you can truly SEE what it is you are looking at.
Sometimes when you are working with volumes it might be more helpful to draw a cross section of the figure, and not necessarily the complete 3D image. This is very common with cones and spheres.
Step 2: Read through the problem a second time, identify all the given and implied material, and where it is located on you diagram.
Keep in mind that many parts of your image will need to be labeled with two pieces of data; a flat amount and a rate of change. A side might be $\textcolor[rgb]{}{5}\textcolor[rgb]{}{m}$ in length and growing at $\textcolor[rgb]{}{2}\frac{\textcolor[rgb]{}{m}}{\textcolor[rgb]{}{s}}$ at a specific moment in time. You will want to label that side with both bits of data. Keep in mind as you do your labeling of rates, that rates of change that are increasing or growing would be positive rates of change, and rates of change that are shrinking or getting smaller would be a negative rate of change. Positive and negatives have real world meanings in these problems, so keep track of them.

Pay attention to the units.

A flat amount that is not changing with have “flat” units: ft, m. No word like “per” in the units.

A rate of change piece will have units like $\frac{\textcolor[rgb]{}{m}}{\textcolor[rgb]{}{s}}$ , meters
per
second, or mph, miles
per
hour. If you find yourself using the word “per”, you are most likely looking at a rate of change.

Rate of Change Pieces

Will usually be accompanied by the word “rate”. “The
rate
at which the volume was changing $\textcolor[rgb]{}{2}\frac{{\textcolor[rgb]{}{m}}^{\textcolor[rgb]{}{3}}}{\textcolor[rgb]{}{s}}$.” Or, “The car was moving at a rate of $\textcolor[rgb]{}{5}\textcolor[rgb]{}{\mathit{mph}}$.”

You will also see words like increasing, decreasing, or other action
“ing”
words. For example, “The radius was increasing at $\textcolor[rgb]{}{5}\frac{\textcolor[rgb]{}{m}}{\textcolor[rgb]{}{s}}$.”

Flat Amount Pieces

The language will generally say a piece of the diagram “is” this. For example, “The radius
is
$\textcolor[rgb]{}{5}\textcolor[rgb]{}{m}$.” Or “The height of the building is $\textcolor[rgb]{}{147}\textcolor[rgb]{}{\mathit{ft}}$.” Notice there is no description of movement.

Additional flat amounts and rates of change sometimes need to be hunted down by you.

They may not give you all the flat amounts that you need to complete your diagram. If they tell you a car has been going $\textcolor[rgb]{}{5}\textcolor[rgb]{}{\mathit{mph}}$ for 2 hours, they are quietly telling you that the car also went 10 miles.

One of the most common unspoken given bits of information is a rate of change value for some portion of the diagram that is zero. Take a look at
all
the parts of your diagram and make sure that they
actually
change
at all in the context of this problem. Sometimes there might be no change occurring along one piece of your diagram, and that is the problem quietly telling you the rate of change is
zero
.
Step 3: Identify the rate of change you are being asked to find.
Are you being asked how the volume is changing, the height is changing, the angle is increasing? What is the rate of change you must hunt down?
Step 4: Find an equation that relates the flat amounts to each other.
You must know your similar triangle relationships, geometry formulas, and your trig formulas.
Step 5: Take the derivative of the equation using implicit differentiation.
Step 6: Plug in all of the pieces you have identified in Step 2 into the derivative equation you just created, and solve for the piece you need.

If all goes well, then you should have only the rate of change that you are trying to find left in the equation as a variable. You then solve for that rate of change variable, and you are done.

If you are still missing more than just the one rate of change you have been asked to find, then you will need to:

Reevaluate the information given to you in the original language of the problem. See if there is an
implied
amount that you were not directly given a value for that you could logic out a value for. For example, if they tell you a car has been going
5mph
for 2 hours, they are quietly telling you that the car also went 10 miles.

See if there is another connection through a
second
equation. See if you can solve that
second
equation for a variable that you can plug into your actual derivative equation. You want to get the derivative equation all talking variables you have values for. For example, you might have a volume formula that relies on the height and the radius, $V=\frac{1}{3}\pi {r}^{2}h$; you have a rate of change for the radius, $\frac{\mathit{dr}}{\mathit{dt}}$, but you don’t have a rate of change for the height, $\frac{\mathit{dh}}{\mathit{dt}}$. That means you need to find a way to relate the
radius
and the
height
in another equation, separate from the one you care about. You can plug a value in for
h
in the volume formula that is
in terms of
r
. Then you will have a volume formula that only uses a
radius
and the derivative of that formula will only need rates of change for a
radius
, $\frac{\mathit{dr}}{\mathit{dt}}$. I know it is a lot to take in, but the more you do of these problems, the less daunting it will feel.

You must know your similar triangle relationships, geometry formulas, and your trig formulas.